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43 lines
1.8 KiB
TeX
43 lines
1.8 KiB
TeX
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\documentclass[11pt,letterpaper]{article}
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\usepackage{natbib}
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\usepackage{graphicx}
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\usepackage[margin=1.in,centering]{geometry}
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\begin{document}
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Consider two lightcurves $x(t)$ and $y(t)$, where $x(t)$ is the driving lightcurve and $y(t)$ is the reprocessed lightcurve. If they are related by a linear impulse response, $g(\tau)$, then:
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\begin{equation}
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y(t) = \int_{-\infty}^{\infty} g(\tau) x(t-\tau) {\rm d}\tau
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\end{equation}
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So, $y(t)$ is a delayed and blurred version of $x(t)$, with the amount of delay and blurring encoded in $g(\tau)$.
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The power spectral density (PSD) of $x(t)$ is calculated from the Fourier transform of $x(t)$, which we denote $X(\nu)$. The PSD is $|X(\nu)|^2 = X^*(\nu)X(\nu)$, where the $^*$ denotes the complex conjugate. From the convolution theorem of Fourier transforms we can write:
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\begin{equation}
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Y(\nu) = G(\nu) X(\nu)
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\end{equation}
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This means it is easy to relate the PSD of the reprocessed lightcurve to the PSD of the driving lightcurve and the impulse response function:
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\begin{equation}
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|Y(\nu)|^2 = |G(\nu)|^2 |X(\nu)|^2
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\end{equation}
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The cross spectrum is defined as
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\begin{equation}
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C(\nu) = X^*(\nu) Y(\nu)
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\end{equation}
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the phase, $\phi$, of which gives the phase lag between X and Y at each Fourier frequency, $\nu$. This can be converted to a time lag through:
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\begin{equation}
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\tau(\nu) = \frac{\phi(\nu)}{2\pi\nu}
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\end{equation}
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Since $Y(\nu) = G(\nu) X(\nu)$, the cross spectrum can be written as:
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\begin{equation}
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C(\nu) = X^*(\nu) G(\nu) X(\nu) = G(\nu) |X(\nu)|^2
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\end{equation}
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thus, for a given impulse response function, one can trivially predict the time lags as a function of frequency, $\tau(\nu)$, by calculating the phase of $G(\nu)$, and the frequency dependence of the lags directly relates to the shape of the response function.
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\end{document}
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